CMSC 471
Machine Learning
Chapter 18.1-18.3, 19.1,
skim 20.4-20.5
Adapted from slides by
Tim Finin and
Marie desJardins.
Some material adopted
from notes by Chuck Dyer
1
Outline
• Machine learning
– What is ML?
– Inductive learning
• Supervised
• Unsupervised
– Decision trees
– Version spaces
– Computational learning theory
2
What is learning?
• “Learning denotes changes in a system that ...
enable a system to do the same task more
efficiently the next time.” –Herbert Simon
• “Learning is constructing or modifying
representations of what is being experienced.”
–Ryszard Michalski
• “Learning is making useful changes in our minds.”
–Marvin Minsky
3
Why learn?
• Understand and improve efficiency of human learning
– Use to improve methods for teaching and tutoring people (e.g., better
computer-aided instruction)
• Discover new things or structure that were previously
unknown to humans
– Examples: data mining, scientific discovery
• Fill in skeletal or incomplete specifications about a domain
– Large, complex AI systems cannot be completely derived by hand
and require dynamic updating to incorporate new information.
– Learning new characteristics expands the domain or expertise and
lessens the “brittleness” of the system
• Build software agents that can adapt to their users or to
other software agents
4
A general model of learning agents
5
Major paradigms of machine learning
• Rote learning – One-to-one mapping from inputs to stored
representation. “Learning by memorization.” Association-based
storage and retrieval.
• Induction – Use specific examples to reach general conclusions
• Clustering – Unsupervised identification of natural groups in data
• Analogy – Determine correspondence between two different
representations
• Discovery – Unsupervised, specific goal not given
• Genetic algorithms – “Evolutionary” search techniques, based on
an analogy to “survival of the fittest”
• Reinforcement – Feedback (positive or negative reward) given at
the end of a sequence of steps
6
The inductive learning problem
• Extrapolate from a given set of examples to make accurate
predictions about future examples
• Supervised versus unsupervised learning
– Learn an unknown function f(X) = Y, where X is an input example
and Y is the desired output.
– Supervised learning implies we are given a training set of (X, Y)
pairs by a “teacher”
– Unsupervised learning means we are only given the Xs and some
(ultimate) feedback function on our performance.
• Concept learning or classification
– Given a set of examples of some concept/class/category, determine if
a given example is an instance of the concept or not
– If it is an instance, we call it a positive example
– If it is not, it is called a negative example
– Or we can make a probabilistic prediction (e.g., using a Bayes net)
7
Supervised concept learning
• Given a training set of positive and negative examples of a
concept
• Construct a description that will accurately classify whether
future examples are positive or negative
• That is, learn some good estimate of function f given a
training set {(x1, y1), (x2, y2), ..., (xn, yn)} where each yi is
either + (positive) or - (negative), or a probability
distribution over +/-
8
Inductive learning framework
• Raw input data from sensors are typically preprocessed to
obtain a feature vector, X, that adequately describes all of the
relevant features for classifying examples
• Each x is a list of (attribute, value) pairs. For example,
X = [Person:Sue, EyeColor:Brown, Age:Young, Sex:Female]
• The number of attributes (a.k.a. features) is fixed (positive,
finite)
• Each attribute has a fixed, finite number of possible values (or
could be continuous)
• Each example can be interpreted as a point in an ndimensional feature space, where n is the number of
attributes
9
Inductive learning as search
• Instance space I defines the language for the training and
test instances
– Typically, but not always, each instance i  I is a feature vector
– Features are also sometimes called attributes or variables
– I: V1 x V2 x … x Vk, i = (v1, v2, …, vk)
• Class variable C gives an instance’s class (to be predicted)
• Model space M defines the possible classifiers
– M: I → C, M = {m1, … mn} (possibly infinite)
– Model space is sometimes, but not always, defined in terms of the
same features as the instance space
• Training data can be used to direct the search for a good
(consistent, complete, simple) hypothesis in the model
space
10
Model spaces
• Decision trees
– Partition the instance space into axis-parallel regions, labeled with class
value
• Version spaces
– Search for necessary (lower-bound) and sufficient (upper-bound) partial
instance descriptions for an instance to be a member of the class
• Nearest-neighbor classifiers
– Partition the instance space into regions defined by the centroid instances
(or cluster of k instances)
•
•
•
•
Associative rules (feature values → class)
First-order logical rules
Bayesian networks (probabilistic dependencies of class on attributes)
Neural networks
11
Model spaces
I
-
I
-
-
+
+
+
+
I
Decision
tree
-
Nearest
neighbor
+
+
Version space
12
Learning decision trees
•Goal: Build a decision tree to classify
examples as positive or negative
instances of a concept using
supervised learning from a training set
•A decision tree is a tree where
– each non-leaf node has associated with it
an attribute (feature)
–each leaf node has associated with it a
classification (+ or -)
–each arc has associated with it one of the
possible values of the attribute at the node
from which the arc is directed
Color
green
Size
big
-
blue
red
+
Shape
small
+
square round
Size
big
-
+
small
+
•Generalization: allow for >2 classes
–e.g., {sell, hold, buy}
13
Decision tree-induced partition –
example
I
Color
green
Size
big
-
blue
red
+
Shape
small
+
square round
Size
big
-
+
small
+
14
Inductive learning and bias
• Suppose that we want to learn a function f(x) = y and we
are given some sample (x,y) pairs, as in figure (a)
• There are several hypotheses we could make about this
function, e.g.: (b), (c) and (d)
• A preference for one over the others reveals the bias of our
learning technique, e.g.:
– prefer piece-wise functions
– prefer a smooth function
– prefer a simple function and treat outliers as noise
15
Preference bias: Ockham’s Razor
• A.k.a. Occam’s Razor, Law of Economy, or Law of
Parsimony
• Principle stated by William of Ockham (1285-1347/49), a
scholastic, that
– “non sunt multiplicanda entia praeter necessitatem”
– or, entities are not to be multiplied beyond necessity
• The simplest consistent explanation is the best
• Therefore, the smallest decision tree that correctly classifies
all of the training examples is best.
• Finding the provably smallest decision tree is NP-hard, so
instead of constructing the absolute smallest tree consistent
with the training examples, construct one that is pretty small
16
R&N’s restaurant domain
• Develop a decision tree to model the decision a patron
makes when deciding whether or not to wait for a table at a
restaurant
• Two classes: wait, leave
• Ten attributes: Alternative available? Bar in restaurant? Is it
Friday? Are we hungry? How full is the restaurant? How
expensive? Is it raining? Do we have a reservation? What
type of restaurant is it? What’s the purported waiting time?
• Training set of 12 examples
• ~ 7000 possible cases
17
A training set
18
A decision tree
from introspection
19
ID3
• A greedy algorithm for decision tree construction developed
by Ross Quinlan, 1987
• Top-down construction of the decision tree by recursively
selecting the “best attribute” to use at the current node in the
tree
– Once the attribute is selected for the current node,
generate children nodes, one for each possible value of
the selected attribute
– Partition the examples using the possible values of this
attribute, and assign these subsets of the examples to the
appropriate child node
– Repeat for each child node until all examples associated
with a node are either all positive or all negative
20
Choosing the best attribute
• The key problem is choosing which attribute to
split a given set of examples
• Some possibilities are:
– Random: Select any attribute at random
– Least-Values: Choose the attribute with the smallest
number of possible values
– Most-Values: Choose the attribute with the largest
number of possible values
– Max-Gain: Choose the attribute that has the largest
expected information gain–i.e., the attribute that will
result in the smallest expected size of the subtrees rooted
at its children
• The ID3 algorithm uses the Max-Gain method of
selecting the best attribute
21
Restaurant example
Patrons versus Type
French
Y
N
Italian
Y
N
Y
NY
Y
NY
Thai N
Burger N
Empty
Some
Full
22
Splitting
examples
by testing
attributes
23
ID3-induced
decision tree
24
Information theory
• If there are n equally probable possible messages, then the
probability p of each is 1/n
• Information conveyed by a message is -log(p) = log(n)
• E.g., if there are 16 messages, then log(16) = 4 and we need 4
bits to identify/send each message
• In general, if we are given a probability distribution
P = (p1, p2, .., pn)
• Then the information conveyed by the distribution (aka entropy
of P) is:
I(P) = -(p1*log(p1) + p2*log(p2) + .. + pn*log(pn))
25
Information theory II
• Information conveyed by distribution (a.k.a. entropy of P):
I(P) = -(p1*log(p1) + p2*log(p2) + .. + pn*log(pn))
• Examples:
– If P is (0.5, 0.5) then I(P) is 1
– If P is (0.67, 0.33) then I(P) is 0.92
– If P is (1, 0) then I(P) is 0
• The more uniform the probability distribution, the greater
its information: More information is conveyed by a message
telling you which event actually occurred
• Entropy is the average number of bits/message needed to
represent a stream of messages
26
The Entropy Function Relative to
Boolean Classification
1.0
Entropy
0.0
Example taken from
Tom Mitchell’s
Machine Learning
0.5
1.0
Proportion of positive examples
27
Huffman code
• In 1952 MIT student David Huffman devised, in the course of
doing a homework assignment, an elegant coding scheme
which is optimal in the case where all symbols’ probabilities are
integral powers of 1/2.
• A Huffman code can be built in the following manner:
– Rank all symbols in order of probability of occurrence
– Successively combine the two symbols of the lowest
probability to form a new composite symbol; eventually we
will build a binary tree where each node is the probability of
all nodes beneath it
– Trace a path to each leaf, noticing the direction at each node
28
Huffman code example
M
Msg.
A
B
C
D
Prob.
.125
.125
.25
.5
A 000 3 0.125 0.375
B 001 3 0.125 0.375
C 01 2 0.250 0.500
D
1 1 0.500 0.500
average message length
1.750
1
0
1
.5
.5
1
0
.25
.25
0
.125
A
C
1
.125
B
code length prob
D
If we use this code to many
messages (A,B,C or D) with this
probability distribution, then, over
time, the average bits/message
should approach 1.75
29
So what does the Huffman code have
to do with information theory?
• Shannon’s entropy gives the average length of the smallest
encoding theoretically possible for a weighted alphabet.
• The information theoretical limit to encoding this alphabet
is:
-0.5*log(0.5) – 0.25*log(0.25) – 0.125*log(0.125) – 0.125*log(0.125) = 1.75
• Huffman’s code is optimal in the information theoretical
sense, and yields encodings which are very very close to the
theoretical limit.
30
Information for classification
• If a set T of records is partitioned into disjoint exhaustive
classes (C1,C2,..,Ck) on the basis of the value of the class
attribute, then the information needed to identify the class of
an element of T is
Info(T) = I(P)
where P is the probability distribution of partition (C1,C2,..,Ck):
P = (|C1|/|T|, |C2|/|T|, ..., |Ck|/|T|)
C1
C1
C3
C2
C2
C3
Low information
High information
31
Information for classification II
• If we partition T w.r.t attribute X into sets {T1,T2, ..,Tn} then
the information needed to identify the class of an element of
T becomes the weighted average of the information needed
to identify the class of an element of Ti, i.e. the weighted
average of Info(Ti):
Info(X,T) =
C1
S|T |/|T| * Info(T )
i
C3
C2
High information
i
C1
C3
C2
Low information
32
Information gain
• Consider the quantity Gain(X,T) defined as
Gain(X,T) = Info(T) - Info(X,T)
• This represents the difference between
– information needed to identify an element of T and
– information needed to identify an element of T after the value of attribute X
has been obtained
That is, this is the gain in information due to attribute X
• We can use this to rank attributes and to build decision trees where at each
node is located the attribute with greatest gain among the attributes not yet
considered in the path from the root
• The intent of this ordering is:
– To create small decision trees so that records can be identified after only a few
questions
– To match a hoped-for minimality of the process represented by the records
being considered (Occam’s Razor)
33
Computing information gain
S|T |/|T| * log(|T |/|T|)
Info(X,T) = S|T |/|T| * Info(T )
I(T) = -
i
French
Y
N
Italian
Y
N
i
i
i
Gain(X,T) = Info(T) - Info(X,T)
I(T) = - (.5 log .5 + .5 log .5)
= .5 + .5 = 1
I (Patrons, T) =
1/6 (0) + 1/3 (0) +
1/2 (- 2/3 log 2/3 – 1/3 log 1/3)
= 1/2 (2/3*.6 + 1/3*1.6)
= .47
Thai
N
Y
NY
Burger
N
Y
NY
Empty
Some
Full
I (Type, T) =
1/6 (1) + 1/6 (1) +
1/3 (1) + 1/3 (1) = 1
Gain (Patrons, T) = 1 - .47 = .53
Gain (Type, T) = 1 – 1 = 0
34
The ID3 algorithm is used to build a decision tree, given a set of non-categorical attributes
C1, C2, .., Cn, the class attribute C, and a training set T of records.
function ID3 (R: a set of input attributes,
C: the class attribute,
S: a training set) returns a decision tree;
begin
If S is empty, return a single node with value Failure;
If every example in S has the same value for C, return
single node with that value;
If R is empty, then return a single node with most
frequent of the values of C found in examples S;
[note: there will be errors, i.e., improperly classified
records];
Let D be attribute with largest Gain(D,S) among attributes in R;
Let {dj| j=1,2, .., m} be the values of attribute D;
Let {Sj| j=1,2, .., m} be the subsets of S consisting
respectively of records with value dj for attribute D;
Return a tree with root labeled D and arcs labeled
d1, d2, .., dm going respectively to the trees
ID3(R-{D},C,S1), ID3(R-{D},C,S2) ,.., ID3(R-{D},C,Sm);
end ID3;
35
How well does it work?
Many case studies have shown that decision trees are
at least as accurate as human experts.
– A study for diagnosing breast cancer had humans
correctly classifying the examples 65% of the
time; the decision tree classified 72% correct
– British Petroleum designed a decision tree for gasoil separation for offshore oil platforms that
replaced an earlier rule-based expert system
– Cessna designed an airplane flight controller using
90,000 examples and 20 attributes per example
36
Extensions of the decision tree
learning algorithm
•
•
•
•
•
•
•
Using gain ratios
Real-valued data
Noisy data and overfitting
Generation of rules
Setting parameters
Cross-validation for experimental validation of performance
C4.5 is an extension of ID3 that accounts for unavailable
values, continuous attribute value ranges, pruning of
decision trees, rule derivation, and so on
37
Using gain ratios
• The information gain criterion favors attributes that have a large
number of values
– If we have an attribute D that has a distinct value for each
record, then Info(D,T) is 0, thus Gain(D,T) is maximal
• To compensate for this Quinlan suggests using the following
ratio instead of Gain:
GainRatio(D,T) = Gain(D,T) / SplitInfo(D,T)
• SplitInfo(D,T) is the information due to the split of T on the
basis of value of categorical attribute D
SplitInfo(D,T) = I(|T1|/|T|, |T2|/|T|, .., |Tm|/|T|)
where {T1, T2, ..., Tm} is the partition of T induced by value of D
38
Computing gain ratio
I(T) = 1
French
Y
N
I (Patrons, T) = .47
Italian
Y
N
I (Type, T) = 1
Gain (Patrons, T) =.53
Gain (Type, T) = 0
Thai
N
Y
NY
Burger
N
Y
NY
Empty
Some
Full
SplitInfo (Patrons, T) = - (1/6 log 1/6 + 1/3 log 1/3 + 1/2 log 1/2)
= 1/6*2.6 + 1/3*1.6 + 1/2*1
= 1.47
SplitInfo (Type, T) = 1/6 log 1/6 + 1/6 log 1/6 + 1/3 log 1/3 + 1/3 log 1/3
= 1/6*2.6 + 1/6*2.6 + 1/3*1.6 + 1/3*1.6 = 1.93
GainRatio (Patrons, T) = Gain (Patrons, T) / SplitInfo(Patrons, T) = .53 / 1.47 = .36
GainRatio (Type, T) = Gain (Type, T) / SplitInfo (Type, T) = 0 / 1.93 = 0
39
Real-valued data
• Select a set of thresholds defining intervals
• Each interval becomes a discrete value of the attribute
• Use some simple heuristics…
– always divide into quartiles
• Use domain knowledge…
– divide age into infant (0-2), toddler (3 - 5), school-aged (5-8)
• Or treat this as another learning problem
– Try a range of ways to discretize the continuous variable and
see which yield “better results” w.r.t. some metric
– E.g., try midpoint between every pair of values
40
Noisy data and overfitting
• Many kinds of “noise” can occur in the examples:
– Two examples have same attribute/value pairs, but different classifications
– Some values of attributes are incorrect because of errors in the data
acquisition process or the preprocessing phase
– The classification is wrong (e.g., + instead of -) because of some error
– Some attributes are irrelevant to the decision-making process, e.g., color of
a die is irrelevant to its outcome
• The last problem, irrelevant attributes, can result in overfitting
the training example data.
– If the hypothesis space has many dimensions because of a large number of
attributes, we may find meaningless regularity in the data that is
irrelevant to the true, important, distinguishing features
– Fix by pruning lower nodes in the decision tree
– For example, if Gain of the best attribute at a node is below a threshold,
stop and make this node a leaf rather than generating children nodes
41
Pruning decision trees
• Pruning of the decision tree is done by replacing a whole
subtree by a leaf node
• The replacement takes place if a decision rule establishes
that the expected error rate in the subtree is greater than in
the single leaf. E.g.,
– Training: one training red true and two training blue falses
– Test: three red falses and one blue true
– Consider replacing this subtree by a single FALSE node.
• After replacement we will have only two errors instead of
four:
Pruned
Test
Training Color
Color
FALSE
red
1 true
0 false
blue
0 true
2 false
red
1 true
3 false
blue
1 true
1 false
2 success
4 failure
42
Converting decision trees to rules
• It is easy to derive a rule set from a decision tree: write a rule
for each path in the decision tree from the root to a leaf
• In that rule the left-hand side is easily built from the label of
the nodes and the labels of the arcs
• The resulting rules set can be simplified:
– Let LHS be the left hand side of a rule
– Let LHS' be obtained from LHS by eliminating some conditions
– We can certainly replace LHS by LHS' in this rule if the subsets of the
training set that satisfy respectively LHS and LHS' are equal
– A rule may be eliminated by using metaconditions such as “if no other
rule applies”
43
Evaluation methodology
• Standard methodology:
1. Collect a large set of examples (all with correct classifications)
2. Randomly divide collection into two disjoint sets: training and test
3. Apply learning algorithm to training set giving hypothesis H
4. Measure performance of H w.r.t. test set
• Important: keep the training and test sets disjoint!
• To study the efficiency and robustness of an algorithm, repeat
steps 2-4 for different training sets and sizes of training sets
• If you improve your algorithm, start again with step 1 to avoid
evolving the algorithm to work well on just this collection
44
Summary: Decision tree learning
• Inducing decision trees is one of the most widely used
learning methods in practice
• Can out-perform human experts in many problems
• Strengths include
–
–
–
–
–
Fast
Simple to implement
Can convert result to a set of easily interpretable rules
Empirically valid in many commercial products
Handles noisy data
• Weaknesses include:
– Univariate splits/partitioning using only one attribute at a time so limits
types of possible trees
– Large decision trees may be hard to understand
– Requires fixed-length feature vectors
– Non-incremental (i.e., batch method)
46
Predicate-Learning Methods
•
•
Decision tree
Version space
Need to provide H
Putting Things Together
with some “structure”
Explicit representation
of hypothesis space H
yes
no
Test
set
Evaluation
Induced
hypothesis h
Training
set 
Learning
procedure L
Object set
Example
set X
Goal predicate
Observable
predicates
Hypothesis
space H
Bias
48
Version Spaces
• The “version space” is the set of all hypotheses that are
consistent with the training instances processed so far.
• An algorithm:
– V := H
;; the version space V is ALL hypotheses H
– For each example e:
• Eliminate any member of V that disagrees with e
• If V is empty, FAIL
– Return V as the set of consistent hypotheses
49
Version Spaces: The Problem
• PROBLEM: V is huge!!
• Suppose you have N attributes, each with k possible values
• Suppose you allow a hypothesis to be any disjunction of
instances
N
• There are kN possible instances  |H| = 2k
• If N=5 and k=2, |H| = 232!!
50
Version Spaces: The Tricks
• First Trick: Don’t allow arbitrary disjunctions
– Organize the feature values into a hierarchy of allowed disjunctions,
e.g.
any-color
dark
black
blue
pale
white
yellow
– Now there are only 7 “abstract values” instead of 16 disjunctive
combinations (e.g., “black or white” isn’t allowed)
• Second Trick: Define a partial ordering on H (“general to
specific”) and only keep track of the upper bound and
lower bound of the version space
• RESULT: An incremental, efficient algorithm!
51
Rewarded Card Example
(r=1) v … v (r=10) v (r=J) v (r=Q) v (r=K)  ANY-RANK(r)
(r=1) v … v (r=10)  NUM(r)
(r=J) v (r=Q) v (r=K)  FACE(r)
(s=) v (s=) v (s=) v (s=)  ANY-SUIT(s)
(s=) v (s=)  BLACK(s)
(s=) v (s=)  RED(s)
A hypothesis is any sentence of the form:
R(r)  S(s)  IN-CLASS([r,s])
where:
• R(r) is ANY-RANK(r), NUM(r), FACE(r), or (r=j)
• S(s) is ANY-SUIT(s), BLACK(s), RED(s), or (s=k)
52
Simplified Representation
For simplicity, we represent a concept by rs, with:
• r  {a, n, f, 1, …, 10, j, q, k}
• s  {a, b, r, , , , }
For example:
• n represents:
NUM(r)  (s=)  IN-CLASS([r,s])
• aa represents:
ANY-RANK(r)  ANY-SUIT(s)  IN-CLASS([r,s])
53
Extension of a Hypothesis
The extension of a hypothesis h is
the set of objects that satisfies h
Examples:
• The extension of f is: {j, q, k}
• The extension of aa is the set of all cards
54
More General/Specific Relation
•
•
Let h1 and h2 be two hypotheses in H
h1 is more general than h2 iff the extension of h1 is a
proper superset of the extension of h2
Examples:
• aa is more general than f
• f is more general than q
• fr and nr are not comparable
55
More General/Specific Relation
Let h1 and h2 be two hypotheses in H
h1 is more general than h2 iff the extension of h1 is a
proper superset of the extension of h2
• The inverse of the “more general” relation is the “more
specific” relation
• The “more general” relation defines a partial ordering on
the hypotheses in H
•
•
56
Example: Subset of Partial Order
aa
na
4a
ab
a
nb
n
4b
4
57
G-Boundary / S-Boundary of V
• A hypothesis in V is most general iff no hypothesis in V is
more general
• G-boundary G of V: Set of most general hypotheses in V
59
G-Boundary / S-Boundary of V
• A hypothesis in V is most general iff no hypothesis in V is
more general
• G-boundary G of V: Set of most general hypotheses in V
• A hypothesis in V is most specific iff no hypothesis in V is
more specific
• S-boundary S of V: Set of most specific hypotheses in V
Image taken from
http://en.wikipedia.org/wiki/Version_space
60
Example: G-/S-Boundaries of V
G
aa
na
ab
4b
n
We replace every hypothesis in S
Now suppose that 4 is
whose
extension
doesa
not
4a
nb
given as a positive example
contain 4 by its generalization set
S
1
…
4
…
k
61
Example: G-/S-Boundaries of V
aa
na G and Sab
Here, both
have size 1.
This is not the case in general!
4a
a
nb
n
4b
4
62
Example: G-/S-Boundaries of V
na
4a
The
aa generalization set
of an hypothesis h is the
set of the
ab hypotheses
that are immediately more
general
than a
h
nb
n
4b
Let 7 be the next
(positive) example
4
Generalization
set of 4
63
Example: G-/S-Boundaries of V
aa
na
4a
ab
a
nb
n
4b
Let 7 be the next
(positive) example
4
64
Example: G-/S-Boundaries of V
Specialization
set of aa
aa
na
ab
a
nb
Let 5 be the next
(negative) example
n
65
Example: G-/S-Boundaries of V
G and S, and all hypotheses in between
form exactly the version space
ab
a
nb
n
66
Example: G-/S-Boundaries of V
At this stage …
ab
No
Yes
Maybe
Do 8, 6, j
satisfy CONCEPT?
a
nb
n
67
Example: G-/S-Boundaries of V
ab
a
nb
Let 2 be the next
(positive) example
n
68
Example: G-/S-Boundaries of V
ab
nb
Let j be the next
(negative) example
69
Example: G-/S-Boundaries of V
+ 4 7 2
– 5 j
nb
NUM(r)  BLACK(s)  IN-CLASS([r,s])
70
Example: G-/S-Boundaries of V
Let us return to the
version space …
… and let 8 be the next
(negative) example
ab
nb
a
The sole most specific
hypothesis disagrees with n
this example, so no
hypothesis in H agrees with
all examples
71
Example: G-/S-Boundaries of V
Let us return to the
version space …
… and let j be the next
(positive) example
ab
nb
a
The only most general
hypothesis disagrees with n
this example, so no
hypothesis in H agrees with
all examples
72
Version Space Update
1. x  new example
2. If x is positive then
(G,S)  POSITIVE-UPDATE(G,S,x)
3. Else
(G,S)  NEGATIVE-UPDATE(G,S,x)
4. If G or S is empty then return failure
73
POSITIVE-UPDATE(G,S,x)
1. Eliminate all hypotheses in G that do not
agree with x
74
POSITIVE-UPDATE(G,S,x)
1. Eliminate all hypotheses in G that do not
agree with x
2. Minimally generalize all hypotheses in S
until they are consistent with x
Using the generalization
sets of the hypotheses
75
POSITIVE-UPDATE(G,S,x)
1. Eliminate all hypotheses in G that do not
agree with x
2. Minimally generalize all hypotheses in S
until they are consistent with x
3. Remove from S every hypothesis that is
neither more specific than nor equal to a
hypothesis in G
This step was not needed in the card example
76
POSITIVE-UPDATE(G,S,x)
1. Eliminate all hypotheses in G that do not
agree with x
2. Minimally generalize all hypotheses in S
until they are consistent with x
3. Remove from S every hypothesis that is
neither more specific than nor equal to a
hypothesis in G
4. Remove from S every hypothesis that is
more general than another hypothesis in S
5. Return (G,S)
77
NEGATIVE-UPDATE(G,S,x)
1. Eliminate all hypotheses in S that do
agree with x
2. Minimally specialize all hypotheses in G
until they are consistent with (exclude) x
3. Remove from G every hypothesis that is
neither more general than nor equal to a
hypothesis in S
4. Remove from G every hypothesis that is
more specific than another hypothesis in G
5. Return (G,S)
78
Example-Selection Strategy
• Suppose that at each step the learning procedure
has the possibility to select the object (card) of the
next example
• Let it pick the object such that, whether the
example is positive or not, it will eliminate onehalf of the remaining hypotheses
• Then a single hypothesis will be isolated in O(log
|H|) steps
79
Example
aa
na
• 9?
• j?
• j?
ab
a
nb
n
80
Example-Selection Strategy
• Suppose that at each step the learning procedure
has the possibility to select the object (card) of the
next example
• Let it pick the object such that, whether the
example is positive or not, it will eliminate onehalf of the remaining hypotheses
• Then a single hypothesis will be isolated in O(log
|H|) steps
• But picking the object that eliminates half the
version space may be expensive
81
Noise
• If some examples are misclassified, the version space may
collapse
• Possible solution:
Maintain several G- and S-boundaries, e.g., consistent with
all examples, all examples but one, etc…
82
VSL vs DTL
• Decision tree learning (DTL) is more efficient if
all examples are given in advance; else, it may
produce successive hypotheses, each poorly
related to the previous one
• Version space learning (VSL) is incremental
• DTL can produce simplified hypotheses that do
not agree with all examples
• DTL has been more widely used in practice
83
Computational learning theory
• Probably approximately correct (PAC) learning:
– Sample complexity (# of examples to “guarantee” correctness)
grows with the size of the model space
• Stationarity assumption: Training set and test sets are drawn
from the same distribution
– Lots of recent work on what to do if this assumption is violated, but
you know something about the relationship between the two
distributions
• Theoretical results apply to fairly simple learning models
(e.g., decision list learning)
84